# Levi Civita symbol on Curl of Vector cross product

1. Oct 3, 2014

### Nikitin

1. The problem statement, all variables and given/known data
Use the LC symbol to calculate the following: $$\nabla \times \frac{\vec{m} \times \hat{r}}{r^2}$$

Where $\vec{m}$ is just a vector, and $\hat{r}$ is the unit radial vector and $r$ is the length of the radial vector.
2. Relevant equations
On the Levi Civita symbol: http://folk.uio.no/patricg/teaching/a112/levi-civita/

3. The attempt at a solution

I've gotten to $\epsilon_{kij} \epsilon_{klm} m_l \frac{r^2 \partial_j r_m - r_m \partial_j r^2}{r^4}$, but I don't know how to take the derivatives..

2. Oct 3, 2014

### Fredrik

Staff Emeritus
That result doesn't look right to me. How did you get it? Note that if $\hat r=\frac{\vec r}{r}$, then
$$\frac{\hat r}{r^2}=\frac{\vec r}{r^3}=\bigg(\frac{r_i}{(r_jr_j)^{\frac 3 2}}\bigg) \hat e_i.$$

3. Oct 3, 2014

### RUber

On the link you provided, there is a proof for $A\times B \times C=B(A \cdot C) - C(A \cdot B)$ using the LC method.
If $\vec{m}$ is just a vector, does that mean constant, i.e. derivatives equal zero?
That should leave you with something like $\vec{m} ( \nabla \cdot \frac{\hat{r}}{r^2})$
You could also look at $\frac{\hat{r}}{r^2}=\frac{1}{x^2+y^2+z^2}\hat{r}$ so taking the partial derivatives should not be too much of a challenge.

4. Oct 4, 2014

### Nikitin

Ruber, yes but here you have a curl along with a function ($r^2$) involved.

But shouldn't it be much easier to perform the derivation in spherical coordinates? This is what's creating the struggle.

I just wrote up the definition, ordered the resulting equation and used the product rule to get started with the derivations. Note, my $r_m$'s are actually the $m$th coordinate of $\hat{r}$, of not $\vec{r}$.

So basically I started with this: $$\epsilon_{ijk} \partial_j \epsilon_{klm} m_l \frac{r_m}{r^2}$$, and used the product rule after ordering to arrive to the equation in the OP.

5. Oct 4, 2014

### Fredrik

Staff Emeritus
But the derivatives are with respect to the mth coordinate of $\vec r$, so with your definition of $r_m$, you don't have $\partial_i x_j=\delta_{ij}$. Instead you have to do something like this (where $x_m$ is the mth component of $\vec r=r\hat r$):
$$\partial_i r_j =\frac{\partial}{\partial x_i} r_j =\frac{\partial}{\partial x_i}\frac{x_j}{\sqrt{x_kx_k}}.$$

6. Oct 4, 2014

### Nikitin

But since $\hat{r}$ is just a one-component vector, shouldn't it be extremely straightforward to use spherical-coordinates $\partial_i$ on it? Why do you change to xyz coordinates?

7. Oct 4, 2014

### vela

Staff Emeritus
It doesn't seem so straightforward to me. In spherical coordinates, $\vec{r}$ depends on all three coordinates in a relatively complicated way whereas in cartesian coordinates you have simply $\partial_i x_j = \delta_{ij}$.

8. Oct 4, 2014

### Fredrik

Staff Emeritus
I never changed coordinates. I just used the definitions. The cross product is defined by $x\times y=(x\times y)_i e_i=(\varepsilon_{ijk}x_jy_k) e_i$, where the $e_i$ are the standard basis vectors, not the basis vectors associated with the spherical coordinate system. $\nabla\times f(x)$ is defined similarly, as $\big(\varepsilon_{ijk}\frac{\partial}{\partial x_j}f_k(x)\big) e_i$.

In this problem, we have
$$f(x)=m\times\frac{\hat r}{r^2}=m\times\frac{\vec r}{r^3},$$ and therefore
$$f_k(x)=\varepsilon_{kln}m_l\frac{x_n}{r^3}=\varepsilon_{kln}m_l\frac{x_n}{(x_px_p)^{\frac 3 2}}.$$ I don't know if there's a way to use both spherical coordinates and the Levi-Civita symbol. (Note that the problem requires you to use the latter). If you know such a way, you can certainly try it.

9. Oct 4, 2014

### Nikitin

OK fine I'll just do it as you said. Thanks.